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Chapter 13: Q. 18 (page 1004)

Explain why it would be difficult to evaluate the double integrals in Exercises 18 and 19 as iterated integrals.

∫∫RexydAwhereR=x,yI1≤x≤2and1≤y≤3

Short Answer

Expert verified

The given double integral is difficult to evaluate using iterated integrals.

Step by step solution

01

Step 1. Given Information.

The iterated integral is,

∫∫RexydAwhereR=x,yI1≤x≤2and1≤y≤3

02

Step 2. Explanation.

Evaluating the first integral with respect to yand treating xas constant.

∫∫RexydA=∫12∫13exydydx=∫12∫13exydydx=∫121xexyy=1y=3dx=∫121xe3x-1xexdx

Now the resultant function 1xe3x-1xexdoes not have a simple anti-derivative to evaluate the next step which is integrating with respect to variablex. The given double integral is difficult to evaluate using iterated integrals.

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Most popular questions from this chapter

Describe the three-dimensional region expressed in each iterated integral in Exercises 35–44.

∫-33∫-9-z29-z2∫03-x2+z2f(x,y,z)dydxdz

Find the masses of the solids described in Exercises 53–56.

The solid bounded above by the hyperboloid with equation z=x2-y2 and bounded below by the square with vertices (2, 2, −4), (2, −2, −4), (−2, −2, −4), and (−2, 2, −4) if the density at each point is proportional to the distance of the point from the plane with equationz = −4.

Evaluate Each of the integrals in exercises 33-36 as an iterated integral and then compare your answer with thoise you found in exercise 29-32

∫∫Rxy3dAWhereR={(x,y)|-2≤x≤2,-1≤y≤1}

Use Definition 13.4to evaluate the double integrals in Exercises 29–32.

localid="1649936867482" ∬xx2ydA

whereR={(x,y)∣1≤x≤3and0≤y≤2}

Use Definition 13.4to evaluate the double integrals in Exercises 29–32.

∬Rxy3dA

whereR={(x,y)∣-2≤x≤2&-1≤y≤1}
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